Curtis (Velocity-Compounded) Impulse Turbine: 25-Nav-B5 Marine Control Systems, May 2016, Q4

Question 4: Curtis (Velocity-Compounded) Impulse Turbine (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. First-stage inlet absolute steam velocity V_s1 = 1411 m/s; nozzle angle θ = 20°; symmetric, frictionless fixed and moving blades; two moving rows (n = 2); steam flow 100 kg/s.

Find. (a) optimum blade speed; (b)–(c) the velocity diagrams and all velocities/angles; (d) work per stage; (e) total power and blade efficiency.

Velocity triangles (blade motion U to the right). Symmetric frictionless blades keep the relative speed constant across each row; the fixed guide row between stages reverses the absolute whirl to feed the second moving row.

Approach. For a velocity-compounded wheel with n moving rows the optimum blade speed is $U=\tfrac{V_{s1}\cos\theta}{2n}$; build the inlet triangle, propagate the constant relative speed through the symmetric rows, and sum the Euler work.

(a) Optimum blade velocity. $U=\dfrac{V_{s1}\cos\theta}{2n}=\dfrac{1411\cos20^{\circ}}{2(2)}=\boxed{331.5\ \text{m/s}}$.
(c) First-stage inlet triangle. Inlet whirl $V_{w1}=V_{s1}\cos20^{\circ}=1326\ \text{m/s}$, flow component $V_{f}=V_{s1}\sin20^{\circ}=482.6\ \text{m/s}$. Relative velocity $W_1=\sqrt{(V_{w1}-U)^2+V_f^2}=\sqrt{994.5^2+482.6^2}=1105\ \text{m/s}$ at blade angle $\beta_1=\tan^{-1}\dfrac{482.6}{994.5}=25.9^{\circ}$.
Stage-1 exit. Symmetric frictionless blade $\Rightarrow |W_2|=|W_1|=1105\ \text{m/s}$ at $\beta_2=25.9^{\circ}$ backwards, so the exit whirl is $V_{w2}=U-994.5=-662.9\ \text{m/s}$ and the absolute exit velocity is $C_2=\sqrt{662.9^2+482.6^2}=820\ \text{m/s}$ at $36.1^{\circ}$ to the plane of rotation (against the blade motion).
Fixed (guide) row and stage-2 inlet. The symmetric fixed blades (inlet and exit angles $36.1^{\circ}$) turn the 820 m/s flow without loss so it re-enters at $36.1^{\circ}$ in the direction of motion: $V_{w3}=+662.9$, $V_f=482.6\ \text{m/s}$. Relative to the second moving row, $W_3=\sqrt{(662.9-331.5)^2+482.6^2}=585\ \text{m/s}$ at $\beta_3=\tan^{-1}(482.6/331.4)=55.5^{\circ}$.
Stage-2 exit. Symmetric blade $\Rightarrow W_4=585\ \text{m/s}$ at $55.5^{\circ}$ backwards, so $V_{w4}=331.5-331.4\approx0$: the steam leaves axially at $C_4=482.6\ \text{m/s}$, which is exactly the minimum-exit-energy condition that defines the optimum blade speed (leaving loss $482.6^2/2=116\ \text{kJ/kg}$).
(d) Work per stage (Euler, $w=U\Delta V_w$). Stage 1 $w_1=659.3\ \text{kJ/kg}$; stage 2 $w_2=219.8\ \text{kJ/kg}$ — the classic $3:1$ split of a two-row Curtis wheel, $\boxed{w_{tot}=879.0\ \text{kJ/kg}}$.
(e) Total power. $P=\dot m\,w_{tot}=100(879.0)=\boxed{87.9\ \text{MW}}$.
(f) Blade efficiency. $\eta_b=\dfrac{w_{tot}}{V_{s1}^2/2}=\dfrac{879.0\times10^3}{1411^2/2}=\boxed{88.3\%}=\cos^{2}20^{\circ}$, the theoretical maximum for a frictionless velocity-compounded stage.

Question 4 — final results
Quantity	Value
(a) Optimum blade speed U	331.5 m/s
(c) W₁ / β₁	1105 m/s / 25.9°
(c) Stage-1 exit absolute C₂ / angle	820 m/s / 36.1°
(c) Stage-2 relative W₃ = W₄ / β₃	585 m/s / 55.5°
(c) Final exit C₄	482.6 m/s (axial)
(d) Work — stage 1 / stage 2	659.3 / 219.8 kJ/kg
(d) Total work	879.0 kJ/kg
(e) Power at 100 kg/s	87.9 MW
(e) Blade efficiency	88.3% (= cos²20°)

Topic. Velocity compounding (Curtis staging) of an impulse steam turbine and its optimum blade-speed ratio.

Key relations. Optimum $U=V_{s1}\cos\theta/(2n)$ for n moving rows; Euler work per row $w=U\,\Delta V_w$; ideal blade efficiency $\eta_b=\cos^{2}\theta$; work split across two rows $\approx 3:1$.

Why this works. A single very high jet velocity (here 1411 m/s) would need an impractically fast wheel to reach U/V = 0.5. Velocity compounding takes the whole velocity drop in one set of nozzles, then extracts it in stages through alternating moving and fixed rows, so the optimum blade speed is reduced by the number of rows. Because each symmetric frictionless row conserves the relative speed and merely turns the flow, the maximum blade efficiency is fixed by the nozzle angle alone, cos²θ.

Common pitfall. Using the single-stage optimum $U=V\cos\theta/2$ instead of dividing by 2n; and mishandling the sign of the whirl when the fixed guide row reverses the flow into the second row.

Source. Dixon & Hall, Turbomachinery, ch. 4 (axial turbines, velocity compounding); Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory, ch. 7.

Question 4 of 8: Curtis (Velocity-Compounded) Impulse Turbine

Question 4: Curtis (Velocity-Compounded) Impulse Turbine (10 marks)